Transport layer

Question

Comments

4 and 300 ?

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Window size is 12 and time is 1 sec

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Yes that is right

I understood congestion window threshold as congestion window size

Explanation :

Threshold is 4 => till 4 we will be in slow start phase after reaching threshold we move to collision avoidance

Slow start => 2 4 ( 2 RTT's required)

Collision Avoidance==> 5 6 7 8 9 10 11 12 ( 8 RTT's required)

=> total 10 RTTs required i.e. 10* 100 = 1000ms 

Answer 1

@Deepalitrapti

A) Receiver can handle maximum 12 segments. So sender can send maximum 12 segments (if we can not consider network's capacity if network capacity is given then sender can send $Min(Receiver's capacity , Network's capacity)$).

B)  Threshold is 4 .. means start with 1 and grow exponentially till 4 , after that grow linearly. 

So it sender send pkt like this 

1  |   2   |  4  |   5  |  6  |   7   |  8  |   9   |   10  |   11   |  12  

So total 10 RRT needs to reaches this window size, then time is 

   $ 10 *  RTT = 10*100ms = 1000ms = 1sec$